# The Decomposition Group

Suppose is a number field that is Galois over with group . Fix a prime lying over .

Definition 14.1.1 (Decomposition group)   The of is the subgroup

(Note: The decomposition group is called the splitting group'' in Swinnerton-Dyer. Everybody I know calls it the decomposition group, so we will too.)

Let denote the residue class field of . In this section we will prove that there is a natural exact sequence

where is the of , and . The most interesting part of the proof is showing that the natural map is surjective.

We will also discuss the structure of and introduce Frobenius elements, which play a crucial roll in understanding Galois representations.

Recall that acts on the set of primes lying over . Thus the decomposition group is the stabilizer in of . The orbit-stabilizer theorem implies that equals the orbit of , which by Theorem 13.2.2 equals the number of primes lying over , so .

Lemma 14.1.2   The decomposition subgroups corresponding to primes lying over a given are all conjugate in .

Proof. We have if and only if . Thus if and only if , so . The lemma now follows because, by Theorem 13.2.2, acts transitively on the set of lying over .

The decomposition group is extremely useful because it allows us to see the extension as a tower of extensions, such that at each step in the tower we understand well the splitting behavior of the primes lying over . Now might be a good time to glance ahead at Figure 14.1.2 on page .

We characterize the fixed field of as follows.

Proposition 14.1.3   The fixed field of

for all

is the smallest subfield such that does not split in  (i.e., ).

Proof. First suppose , and note that by Galois theory , and by Theorem 13.2.2, the group acts transitively on the primes of lying over . One of these primes is , and fixes by definition, so there is only one prime of lying over , i.e., does not split in . Conversely, if is such that does not split in , then fixes (since it is the only prime over ), so , hence .

Thus does not split in going from to --it does some combination of ramifying and staying inert. To fill in more of the picture, the following proposition asserts that splits completely and does not ramify in .

Proposition 14.1.4   Let for our fixed prime and Galois extension . Let be for and . Then and , i.e., does not ramify and splits completely in . Also and .

Proof. As mentioned right after Definition 14.1.1, the orbit-stabilizer theorem implies that , and by Galois theory . Thus

Now and , so we must have and . Since and , the proposition follows.

Subsections
William Stein 2004-05-06